2 Answers
2

The formula you ask about is reasonably common. E.g. in simplicial homology/cohomology, we triangulate a space into simplices. The chain complex
made out of these simplices then computes the simplicial homology.

We can also define simplicial cochains: this is just the dual complex to the simplicial chain complex, so concretely a simplicial cochain just attaches a number to each simplex in our space.

If $\Omega$ is a simplex, and $\omega$ is a simplicial cochain, then by definition of the coboundary operator $d$ on simplicial cochains, one has the formula
$$\langle \Omega, d\omega \rangle = \langle \partial \Omega, \omega\rangle.$$
This is more tautological than the de Rham cohomology case, though, because
one doesn't have an a priori notion of simplicial cochains, or of the coboundary operator on them, so the whole theory of simplicial cohomology is defined to make this formula be true. (An entirely analogous story is true if we replace simplicial by singular everywhere in the above.)

What is somewhat special in the de Rham case is that we have an a priori notion both of simplices or other submanifold-like subobjects of a manifold and their boundaries, and of differential forms and the exterior derivative on forms. The formula relating them is in then not a formality; indeed, it allows you to compare simplicial or singular cohomology with de Rham cohomology, and is really the most important ingredient in the the fact that those cohomologies are isomorphic to de Rham cohomology.

If one keeps in the purely simplicial (or singular) context, at first the formula looks completely formal, and indeed simplicial (or singular) cohomology
itself at first seems quite formal (and you can wonder why you need it, when
you already have singular/simplicial homology). But there are other descriptions
of singular cohomology, e.g. via obstruction theory. For example, one has
the isomorphism
$$H^1(X,\mathbb Z) = \text{ homotopy classes of maps } X \to S^1.$$
From this point of view simplicial or singular cohomology classes can take on
a non-formal appearance, and in proving such statements, the formula you ask about plays a key role.

When people say cohomology theory, I usually think of some spectrum. The reason for that is Eilenberg and Steenrod developed axioms for what a cohomology theory on the category of pairs of "nice" spaces should do. They then showed that singular cohomology satisfied these and was essentially unique. De Rham cohomology is sort of miraculous in that it is easy to see the geometry. It is staring you right in the face, these things are forms on your manifold!

The pairing you see in stokes theorem is sort of a really special case for 2 reasons. The first is the inherent geometry in De Rham cohomology. We have a geometric interpretation of what it means to be a De Rham cocycle. And by geometric I mean that for every $\alpha \in H^n_{dR} X$ where $X$ is a manifold, we understand there is some intrinsic structure on $X$ that this is detecting. This is usually not the case for other theories. We have this in $K$-theory with vector bundles and cobordism with families of manifolds, but that is it. It would be a huge deal if someone understood what $\alpha \in tmf^*X$ was supposed to be on $X$ they would win a prize (like a hug or a job or something).

The second is that singular cohomology is one of the few cohomology theories that has a formulation as the homology of some chain complex. You could ask for homology theories on spaces to really be invariants that land in chain complexes and aren't all that well defined. I am pretty sure you are just left with singular cohomology.

So let me be liberal and say that a cohomology theory should be something that satisfies the Eilenberg-Steenrod axioms, except for the dimension axiom. Lets also suppose this cohomology theory has some sort of product, then Brown representability tells us we have a ring spectrum. This is what I mean by a represented cohomology theory. So in this setting $E^*X=[X,E]$ is just the homotopy classes of maps from $X$ into the spectrum $E$, whatever that means. Similarly $E_* X$ is just homotopy classes of maps from the sphere spectrum into $X \wedge E$.

While stokes theorem is really about De Rham theory, it is really foreshadowing.
Integration of differential forms against submanifolds is really a nice model of something called the cap product, which is really only well-defined because of Stokes theorem. The cap product is like a pairing between $E^*X$ and $E_*X$. The cap product is a little more subtle, so I won't talk about it (it's not hard, just a little involved).

To answer your very specific question, there is only such a formula in singular cohomology theories because the boundary operator only makes sense in that setting.

My suggestion is that you try to pair $a \in E^*X$ and $b \in E_*X$ to get an element in $\pi_*E$ (in the case you mention $E=H\mathbb{R}$ the Eilenberg-Maclane spectrum of the reals, which represents de Rham cohomology, and the output would be a real number ... an element of $\pi_0 H \mathbb{R}$).

Let me know if you need help, but remember to use the fact that we are working with represented cohomology and homology theories and that $E$ is a ring spectrum, so it has a product map $\mu: E \wedge E \to E$.

Well I don't understand most of this notation. Also I think there is a typo in each of the three parts of the answer.
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NikolajKJan 13 '12 at 23:06

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Typo as in grammatical or mathematical? If so, please point out my error. also, when I get more time I will point to a better reference or try to flesh it out. You might want to look at the following: math.stackexchange.com/a/96821/627
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Sean TilsonJan 17 '12 at 4:14

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The question is about a Stokes'-type theorem, not just about the existence of a pairing. I agree that the question itself doesn't make much sense though, at least as it's currently phrased.
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Aaron Mazel-GeeJan 17 '12 at 12:09

ah, I see what you mean. Maybe I will edit later, but I would interpret the stokes theorem as a result saying that there is a well defined pairing in terms of chains and cochains. This is the same as saying that the pairing I assert is exists and is well defined up to homotopy. At least that is what it seems like.
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Sean TilsonJan 17 '12 at 23:17

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There should be a ceremony for the Tilson Hug right there together with the Fields medal in IMU meetings!
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Mariano Suárez-Alvarez♦Jan 18 '12 at 19:20